Alternate Segment Theorem (Tangent-Chord): Class 10

Quick answer: The alternate segment theorem says: the angle between a tangent and a chord drawn from the point of contact equals the angle that the chord makes at any point of the circle in the alternate (opposite) segment. If tangent XY touches the circle at A and AB is a chord, then $\angle BAY = \angle ACB$ for any point C on the arc on the other side of AB. It is proved by drawing the diameter through A.

Syllabus note: This theorem is not part of the NCERT Class 10 Circles chapter, which covers only the tangent being perpendicular to the radius and the two tangents from an external point being equal. It is useful for olympiads and other boards. For CBSE answers, use the syllabus theorems, as shown in the second proof of the rider below.

Statement

A tangent XY touches a circle at A, and AB is a chord. The chord divides the circle into two segments.

$\angle BAY$ is the angle between the tangent and the chord.
C is any point on the arc in the segment on the other side of AB from $\angle BAY$ .

Then:

$\angle BAY = \angle ACB$

Similarly, the angle on the other side, $\angle BAX$ , equals the angle that AB subtends at any point of the other arc.

Proof using a diameter

Given: A circle with centre O, a tangent XY at A, a chord AB, and a point C in the alternate segment.

To prove: $\angle BAY = \angle ACB$ .

Construction: Draw the diameter AD through O, with D on the same arc as C. Join BD.

Step 1: The radius OA is perpendicular to the tangent, so $\angle DAY = 90^\circ$ . Therefore

$\angle BAY = 90^\circ - \angle DAB$

Step 2: $\angle ABD$ is an angle in a semicircle, so $\angle ABD = 90^\circ$ .

Step 3: In triangle ABD, the angles add up to $180^\circ$ :

$\angle ADB = 90^\circ - \angle DAB$

Step 4: From Steps 1 and 3, $\angle BAY = \angle ADB$ .

Step 5: D and C lie in the same segment, so the angles subtended by chord AB at D and C are equal: $\angle ADB = \angle ACB$ .

Conclusion: $\angle BAY = \angle ACB$ . Hence proved.

Steps 2 and 5 use "angle in a semicircle" and "angles in the same segment", which are Class 9 circle theorems.

Rider: tangent at the vertex of an isosceles triangle

Triangle ABC is inscribed in a circle with $AB = AC$ . PAQ is the tangent at A. Prove that $PQ \parallel BC$ .

Proof 1: with the alternate segment theorem

Step 1: Since $AB = AC$ , $\angle ABC = \angle ACB$ .

Step 2: By the alternate segment theorem, the angle between tangent AP and chord AB equals the angle in the alternate segment: $\angle PAB = \angle ACB$ .

Step 3: So $\angle PAB = \angle ABC$ . These are alternate angles for the lines PQ and BC with transversal AB.

Conclusion: $PQ \parallel BC$ .

Proof 2: using only Class 10 syllabus theorems

Step 1: Join OA, where O is the centre. The tangent is perpendicular to the radius, so $OA \perp PQ$ .

Step 2: A is equidistant from B and C (since $AB = AC$ ), and so is O (since $OB = OC$ ). So the line AO is the perpendicular bisector of BC, and $OA \perp BC$ .

Step 3: PQ and BC are both perpendicular to the same line OA.

Conclusion: $PQ \parallel BC$ . Hence proved.

Worked example

A tangent at A makes an angle of $65^\circ$ with the chord AB. Find the angle subtended by AB at a point C on the major arc.

Step 1: By the alternate segment theorem, the required angle equals the angle between the tangent and the chord.

Answer: $\angle ACB = 65^\circ$ . A point on the minor arc would see the chord at $180^\circ - 65^\circ = 115^\circ$ , because opposite angles of a cyclic quadrilateral add up to $180^\circ$ .

Summary table

Result	Reason
Tangent ⟂ radius at the point of contact	NCERT Class 10 theorem
Angle in a semicircle is $90^\circ$	Class 9 circle theorem
Angles in the same segment are equal	Class 9 circle theorem
Tangent-chord angle = angle in the alternate segment	Alternate segment theorem (beyond NCERT)
Tangent at the vertex of an isosceles triangle ∥ base	Both proofs above

Exam tips

In CBSE exams, prefer proofs that use the tangent-radius theorem; the alternate segment theorem is not in the textbook.
State the reason for every step, for example "angle in a semicircle".
Mark the alternate segment clearly on your figure before writing the proof.

Common mistakes

Using the angle in the same segment as the tangent-chord angle instead of the alternate one.
Forgetting that the point used in the proof (D) must lie in the same segment as C.
Quoting the alternate segment theorem in a CBSE board answer without proving it.
Confusing alternate angles with corresponding angles when proving lines parallel.

Frequently Asked Questions

What does the alternate segment theorem state?

The angle between a tangent and a chord through the point of contact equals the angle the chord subtends at any point in the alternate segment of the circle.

Is the alternate segment theorem in the CBSE Class 10 syllabus?

No. The NCERT Class 10 Circles chapter covers only the tangent-radius theorem and the equal tangents theorem.

How is the alternate segment theorem proved?

By drawing the diameter through the point of contact and using the tangent-radius theorem, the angle in a semicircle and angles in the same segment.

Why is the tangent at the vertex of an isosceles triangle parallel to the base?

Both the tangent and the base are perpendicular to the line from the vertex through the centre, so they are parallel.

What is the alternate segment?

It is the segment of the circle on the other side of the chord from the angle formed by the tangent and the chord.